mod of a cap - b cap = ¡Ì2 then calculate the value of mod of a cap + ¡Ì3b acp
To solve the given equation, we have:
|a| - |b| = √2
First, let's consider the case when a ≥ 0 and b ≥ 0:
In this case, the equation becomes:
a - b = √2
Next, let's consider the case when a ≥ 0 and b < 0:
In this case, the equation becomes:
a - (-b) = √2 (since the magnitude of a negative number is equal to its positive counterpart)
a + b = √2
Now, let's consider the case when a < 0 and b ≥ 0:
In this case, the equation becomes:
(-a) - b = √2 (since the magnitude of a negative number is equal to its positive counterpart)
-a - b = √2
Finally, let's consider the case when a < 0 and b < 0:
In this case, the equation becomes:
(-a) - (-b) = √2
-b + a = √2
Now, the value of |a| + √3|b| can be calculated in each of the above cases.
- When a ≥ 0 and b ≥ 0:
|a| + √3|b| = a + √3b
- When a ≥ 0 and b < 0:
|a| + √3|b| = a + √3(-b) = a - √3b
- When a < 0 and b ≥ 0:
|a| + √3|b| = (-a) + √3b = -a + √3b
- When a < 0 and b < 0:
|a| + √3|b| = (-a) + √3(-b) = -a - √3b
Therefore, the value of |a| + √3|b| depends on the cases discussed above.
To calculate the value of |â + √3b̂|, we need to find the values of â and b̂ first. Given the equation:
|â - b̂| = √2
We square both sides of the equation to eliminate the square root:
(â - b̂)² = (√2)²
â² - 2âb̂ + b̂² = 2
Now, let's multiply the equation by 3:
3(â² - 2âb̂ + b̂²) = 3(2)
3â² - 6âb̂ + 3b̂² = 6
Now, we are going to work with the expression |â + √3b̂|. We square both sides of the equation:
(|â + √3b̂|)² = (3â² - 6âb̂ + 3b̂²)
Since the square of any value is always positive, we can remove the absolute value sign:
(â + √3b̂)² = 3â² - 6âb̂ + 3b̂²
Now, let's substitute the value of the right side of the equation with the equivalent expression from the previous equation:
(â + √3b̂)² = 6
Taking the square root of both sides, we have:
√(â + √3b̂)² = √6
(â + √3b̂) = ±√6
Therefore, the value of |â + √3b̂| is ±√6.